Question

Find distance between final position of ball and its initial pt. of projection (53° with horizontal) after hitting the wall.
Velocity of ball is 50 m/s.

Dont copy from Google ​

Answer 1

$\huge\orange{\underline{ \overline{ \boxed{ \mathbb\blue{ANSWER}}}}}$

$\sf{ \underline{ \underline{\green{GIVEN :}}}}$

• Angle of projection = 53°
• Velocity of the ball (u) = 50m/sec.
• Coefficient of restitution (e) = $\frac{1}{2}$
• Horizontal distance = 60 m [May be by mistake not mentioned in the question but it is there.]

$\sf{ \underline{ \underline{\green{TO \: FIND :}}}}$

The distance between initial and final position of the ball after hitting the wall.

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$\sf{ \underline{ \underline{\green{SOLUTION :}}}}$

♡ Diagram refer to the Attachment.♡

• Let the point of projection of the ball be 'A' and the final position after hitting the wall be 'B'.
• Let 'C' be the point on the wall where the ball strikes.
• Let 'C' be the point on the wall where the ball strikes. And let 'x' be the distance between initial and final positions i.e 'A' and 'B'.

☆ Now, If we analyse the path traveled by the ball we can find its in the shape of parabolic path. So, the ball actually did a projectile motion.

Now, we need to find the horizontal and vertical speed of the projectile motion.

Here,

Horizontal velocity,$\sf{\:\: U_x = u \cos \theta}$

$\sf{ = 50 (\cos53 \degree)}$

$\sf{ = 50 \times ( \frac{3}{5} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: [{ \cos35 \degree = \frac{3}{5}}}]$

$\sf{ = 30m/s }$

$\boxed{\sf{ \red{\therefore \: U_x = 30m/sec}}}$

Now for Vertical velocity,

Vertical velocity,$\sf{ \:\: U_y = u \sin \theta}$

$\sf{ = 50( \sin53 \degree)}$

$\sf{ = 50 \times ( \frac{4}{5} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: [{ \sin35 \degree = \frac{4}{5}}}]$

$\sf{ = 30m/s }$

$\boxed{\sf{\red{\therefore \: U_y = 40m/sec}}}$

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Now, as the ball hits the wall at point 'C' and returns to the position 'B', so its line of impact is along with the horizontal axis.

And we know that in a projectile motion after hitting any surface the velocity becomes 'e' times the previous velocity.

As, here the line of impact is along horizontal axis, so the new velocity (let say 'v') of the ball while returning from 'C' to 'B' will be,

$\implies \sf{ v \: = e \times U_x}$

[On putting the values]

$\implies \sf{ v \: = \frac{1}{2}\times 30}$

$\boxed{\therefore {\red{\sf{ v \: = 15 m/sec.}}}}$

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Now, Let the time of flight of the ball be 'T'

$\sf{ \implies T \: = \frac{2 u\sin \theta}{g}}$

$\sf{\implies T \: = \frac{2 U_y}{g}}$

$\sf{\implies T \: = \frac{2 \times 40}{10}}$

[Taken g = 10m/s^2]

$\boxed{\sf {\red{ T \: = 8 \: sec}}} \:\:\:\:\:\: \longrightarrow (1)$

Now, let $t_1$ be the time required for forward journey i.e from 'A' to 'C' and $t_2$ be for backward journey i.e from 'C' to 'B'

For Forward Journey,

$\sf{\implies \: t_1 = \frac{Horizontal \: distance}{Horizontal \: Speed}}$

$\sf{\implies \: t_1 = \frac{60}{U_x}}$

$\sf{\implies \: t_1 = \frac{60}{30}}$

$\boxed{\sf{\red {\therefore \: t_1 = 2 \: sec.}}}$

Now,For Backword journey time taken = (Time of flight) - (Time taken for forward journey. )

$\sf{\implies \: t_2 = 8 - 2}\:\:\:\:\:\:\:\:[from (1)]$

$\boxed{\sf{\red{\therefore \: t_2 = 6 \: sec.}}}$

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Now, let us find out the distance travelled from 'A' to 'B'

Distance travelled = Speed(v) × time(t2)

$\sf{ \implies \: x =v \times t_2}$

$\sf{ \implies \: x =15 \times 6}$

$\boxed{\sf{\blue{ \therefore \: x = 90 \:m}}}$

Hence, the distance between initial and final positions is 90m.

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